2cos квадрат x-5cos x-3=0

[latex]\displaystyle 2cos^2x-5cosx-3=0 \;\big| cosx=t\\\\2t^2-5t-3=0\\\\D:25+24=1\\\\t_1,_2=\frac{5\pm1}{4}; \quad x_1=\frac{3}{2}, \quad x_2=1;\\\\1. \; cosx \neq \frac{3}{2}, \; cosx \in [-1;1];\\\\2. \; cosx=1\\x=2\pi n, \; n\in Z.[/latex]